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Math Help - help understanding this notation

  1. #1
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    help understanding this notation

    My professor wants an example of the Schroeder-Bernstein Theorem (If the card A < card B and card B < card A, then card A = card B).

    He wants us to use A=B=\Re, and f:A\longrightarrow B to be f(x)=e^x and g:B\longrightarrow A to be g(x)=\arctan x.

    There is a proof sketch in the book, and he wants us to work through that using our example functions and sets. I'm just confused about the notation here:

    Define A_0=A and B_0=B; A_1=f(A_0) and B_1=g(B_0). Then for each n\in N define:
    A_{2n}=g(A_{2n-1}, A_{2n+1}=f(A_{2n}),
     B_{2n}=f(B_{2n-1}, B_{2n+1}=g(B_{2n}).

    How do I put A=B=\Re into this? I want to say that they're all just \Re but I'm not sure. Help!

    Thank you!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sfitz View Post
    My professor wants an example of the Schroeder-Bernstein Theorem (If the card A < card B and card B < card A, then card A = card B).

    He wants us to use A=B=\Re, and f:A\longrightarrow B to be f(x)=e^x and g:B\longrightarrow A to be g(x)=\arctan x.

    There is a proof sketch in the book, and he wants us to work through that using our example functions and sets. I'm just confused about the notation here:

    Define A_0=A and B_0=B; A_1=f(A_0) and B_1=g(B_0). Then for each n\in N define:
    A_{2n}=g(A_{2n-1}, A_{2n+1}=f(A_{2n}),
     B_{2n}=f(B_{2n-1}, B_{2n+1}=g(B_{2n}).

    How do I put A=B=\Re into this? I want to say that they're all just \Re but I'm not sure. Help!

    Thank you!!

    That is what it says, you then need to observe that A_1=f(A)=f(\Re)=(0,\infty), and B_1=g(B)=g(\Re)=(-\pi/2,\pi/2), and away you go.

    RonL
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